Question

n=3ˏ3 and 5i are zerosˏf(−1)=−312

Answer 1

So far, you know that \(f(x)\) has the following partial form:
\[f(x)=\cdots(x-3)^2(x-5i)\cdots\]
(I'm assuming you do in fact mean to say that 3 is a double root for \(f\); please correct me if it's not.)

The thing about complex roots to a polynomial with real coefficients (another assumption I'm making) is that they always appear in conjugate pairs, which means \(-5i\) is also a root, which means
\[f(x)=\cdots(x-3)^2(x-5i)(x+5i)\]
We're also missing a possible constant multiplier, call it \(a\):
\[f(x)=a(x-3)^2(x-5i)(x+5i)\]
Plug in the given values to solve for \(a\):
\[-312=a(-1-3)^2(-1-5i)(-1+5i)~~\Rightarrow~~a=-\frac{3}{4}\]